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Question Number 18827 by aplus last updated on 30/Jul/17

Commented by aplus last updated on 30/Jul/17

help guys

$$\mathrm{help}\:\mathrm{guys} \\ $$

Commented by mrW1 last updated on 30/Jul/17

no solution

$$\mathrm{no}\:\mathrm{solution} \\ $$

Commented by dioph last updated on 31/Jul/17

if both x and y are > 1: x^x > x, y^y > y ⇒ x^x + y^y > x + y ⇒ 6 > 35, absurd if exactly one from x or y is < 1, say x < 1: x + y = 35 ⇒ y > 34 ⇒ y^y > 34^(34) ⇒ x^x + y^y > 6, absurd if both x and y are < 1: x + y < 2 < 35, absurd

$$\mathrm{if}\:\mathrm{both}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:>\:\mathrm{1}: \\ $$$${x}^{{x}} \:>\:{x},\:{y}^{{y}} \:>\:{y}\:\Rightarrow\:{x}^{{x}} \:+\:{y}^{{y}} \:>\:{x}\:+\:{y} \\ $$$$\Rightarrow\:\mathrm{6}\:>\:\mathrm{35},\:\mathrm{absurd} \\ $$$$\mathrm{if}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{from}\:{x}\:\mathrm{or}\:{y}\:\mathrm{is}\:<\:\mathrm{1}, \\ $$$$\mathrm{say}\:{x}\:<\:\mathrm{1}: \\ $$$${x}\:+\:{y}\:=\:\mathrm{35}\:\Rightarrow\:{y}\:>\:\mathrm{34}\:\Rightarrow\:{y}^{{y}} \:>\:\mathrm{34}^{\mathrm{34}} \\ $$$$\Rightarrow\:{x}^{{x}} \:+\:{y}^{{y}} \:>\:\mathrm{6},\:\mathrm{absurd} \\ $$$$\mathrm{if}\:\mathrm{both}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:<\:\mathrm{1}: \\ $$$${x}\:+\:{y}\:<\:\mathrm{2}\:<\:\mathrm{35},\:\mathrm{absurd} \\ $$

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